Solving Quadratic Equations | 解二次方程

📚 Solving Quadratic Equations | 解二次方程

Quadratic equations are one of the most important topics in IGCSE Mathematics. From factorisation to the quadratic formula, mastering this topic unlocks success in algebra, graphs, and problem-solving across the entire syllabus.

二次方程是 IGCSE 数学中最重要的课题之一。从因式分解到求根公式，掌握这一课题将为你在代数、函数图像以及应用题中的成功奠定坚实基础。

1. What Is a Quadratic Equation? | 什么是二次方程？

A quadratic equation is an equation of the form ax² + bx + c = 0, where a, b and c are constants and a ≠ 0. The highest power of the variable x is 2, which gives the equation its name ‘quadratic’.

二次方程是形如 ax² + bx + c = 0 的方程，其中 a、b、c 为常数，且 a ≠ 0。变量 x 的最高次数为 2，这正是 “二次” 名称的由来。

The coefficient a cannot be zero, because if a = 0, the equation would become linear (bx + c = 0), not quadratic. The values of a, b and c can be positive, negative, or zero, but a must always be non-zero.

系数 a 不能为零，因为若 a = 0，方程将退化为一次方程（bx + c = 0），而不再是二次方程。a、b、c 的值可以为正、为负或为零，但 a 必须始终非零。

ax² + bx + c = 0（a ≠ 0）

2. Solving by Factorisation | 用因式分解法求解

Factorisation is often the quickest method for solving quadratic equations when the expression can be neatly factored. The key principle is the zero product property: if the product of two factors equals zero, then at least one of the factors must be zero.

当二次表达式可以被整洁地分解时，因式分解通常是最快捷的求解方法。其核心原理是零乘积性质：若两个因式的乘积为零，则至少有一个因式必须为零。

To solve x² + 5x + 6 = 0, we first find two numbers that multiply to give 6 and add to give 5. Those numbers are 2 and 3, so we write (x + 2)(x + 3) = 0. Then x + 2 = 0 or x + 3 = 0, giving x = -2 or x = -3.

要解 x² + 5x + 6 = 0，我们首先找到两个数，它们相乘得到 6，相加得到 5。这两个数就是 2 和 3，于是写成 (x + 2)(x + 3) = 0。然后 x + 2 = 0 或 x + 3 = 0，得到 x = -2 或 x = -3。

x² + 5x + 6 = 0 → (x + 2)(x + 3) = 0 → x = -2 或 x = -3

When the coefficient of x² is not 1, factorisation requires a bit more care. For example, to solve 2x² – 5x – 3 = 0, we look for factors of 2 × (-3) = -6 that add to -5. Those are -6 and 1. We rewrite the middle term: 2x² – 6x + x – 3 = 0, then factor by grouping: 2x(x – 3) + 1(x – 3) = 0, giving (2x + 1)(x – 3) = 0. Hence x = -½ or x = 3.

当 x² 的系数不为 1 时，因式分解需要更加小心。例如，要解 2x² – 5x – 3 = 0，我们寻找 2 × (-3) = -6 的因子对，使其和为 -5。这两个数是 -6 和 1。我们改写中间项：2x² – 6x + x – 3 = 0，然后分组分解：2x(x – 3) + 1(x – 3) = 0，得到 (2x + 1)(x – 3) = 0。因此 x = -½ 或 x = 3。

3. Solving by Using the Quadratic Formula | 用求根公式求解

Not every quadratic expression can be factored easily. In such cases, the quadratic formula provides a reliable, universal method. The formula is derived from completing the square and works for any quadratic equation.

并非每个二次表达式都能轻松因式分解。在这种情况下，求根公式提供了一种可靠且通用的方法。该公式由配方法推导而来，适用于任何二次方程。

x = (-b ± √(b² – 4ac)) / 2a

To use the formula, identify the values of a, b and c from the equation, substitute them into the formula, and simplify carefully. The ± symbol indicates that there are usually two solutions: one using the plus sign and one using the minus sign.

使用公式时，先从方程中确定 a、b、c 的值，将其代入公式，然后仔细化简。± 符号表示通常有两个解：一个用加号，一个用减号。

For example, solve x² – 4x – 7 = 0. Here a = 1, b = -4, c = -7. Substituting gives x = (4 ± √(16 + 28)) / 2 = (4 ± √44) / 2 = (4 ± 2√11) / 2 = 2 ± √11. So the solutions are x = 2 + √11 ≈ 5.32 and x = 2 – √11 ≈ -1.32.

例如，解 x² – 4x – 7 = 0。这里 a = 1，b = -4，c = -7。代入得 x = (4 ± √(16 + 28)) / 2 = (4 ± √44) / 2 = (4 ± 2√11) / 2 = 2 ± √11。因此解为 x = 2 + √11 ≈ 5.32 和 x = 2 – √11 ≈ -1.32。

4. Solving by Completing the Square | 用配方法求解

Completing the square is a powerful algebraic technique that transforms a quadratic expression into the form (x + p)² + q. This method is essential not only for solving equations but also for finding the turning point of a quadratic graph.

配方法是一种强大的代数技巧，它将二次表达式转化为 (x + p)² + q 的形式。这种方法不仅对求解方程至关重要，而且对寻找二次函数图像的顶点也必不可少。

To complete the square for x² + 6x + 5, take half of 6 (which is 3), square it (giving 9), and rewrite: x² + 6x + 9 – 9 + 5 = (x + 3)² – 4. The equation (x + 3)² – 4 = 0 then gives (x + 3)² = 4, so x + 3 = ±2, hence x = -1 or x = -5.

要对 x² + 6x + 5 配方，取 6 的一半（即 3），平方得 9，然后改写：x² + 6x + 9 – 9 + 5 = (x + 3)² – 4。方程 (x + 3)² – 4 = 0 得到 (x + 3)² = 4，因此 x + 3 = ±2，从而 x = -1 或 x = -5。

x² + 6x + 5 = (x + 3)² – 4 = 0 → (x + 3)² = 4 → x = -1 或 -5

When the coefficient of x² is not 1, first factor out that coefficient from the x² and x terms, then complete the square inside the brackets. This method also reveals the minimum or maximum point of the parabola, which is ((-p), q) when written as a(x + p)² + q.

当 x² 系数不为 1 时，先从 x² 项和 x 项中提出该系数，再在括号内配方。这种方法还能揭示抛物线的最小值或最大值点：当表达式写作 a(x + p)² + q 时，顶点为 ((-p), q)。

5. The Discriminant: How Many Solutions? | 判别式：有多少个解？

The expression b² – 4ac inside the square root of the quadratic formula is called the discriminant. It tells us how many real solutions a quadratic equation has without actually solving it.

求根公式根号内的表达式 b² – 4ac 称为判别式。它可以在不实际求解的情况下告诉我们二次方程有多少个实数解。

If b² – 4ac > 0, the equation has two distinct real roots.

若 b² – 4ac > 0，方程有两个不相等的实数根。
If b² – 4ac = 0, the equation has exactly one real root (a repeated root).

若 b² – 4ac = 0，方程恰好有一个实数根（重根）。
If b² – 4ac < 0, the equation has no real roots (the roots are complex).

若 b² – 4ac < 0，方程没有实数根（根为复数）。

Geometrically, the discriminant tells us how many times the parabola intersects the x-axis: two points, one point, or no points at all. This connection between algebra and graphs is frequently tested in IGCSE examination questions.

从几何角度看，判别式告诉我们抛物线与 x 轴的交点个数：两个交点、一个交点或没有交点。这种代数与图像之间的联系是 IGCSE 考试中经常考查的内容。

6. Solving Quadratic Equations by Graphical Methods | 用图像法解二次方程

Graphing a quadratic function y = ax² + bx + c produces a parabola. The solutions to the equation ax² + bx + c = 0 correspond exactly to the x-coordinates where the parabola crosses the x-axis.

绘制二次函数 y = ax² + bx + c 的图像会得到一条抛物线。方程 ax² + bx + c = 0 的解恰好对应抛物线与 x 轴交点处的 x 坐标。

To solve graphically, plot the curve carefully for a given range of x values, then read off the x-intercepts. If the parabola touches the x-axis at one point, there is one repeated root; if it never touches the x-axis, there are no real roots.

要用图像法求解，在给定的 x 值范围内仔细绘制曲线，然后读取与 x 轴的交点。如果抛物线在 x 轴上只有一个切点，则有一个重根；如果抛物线始终不与 x 轴相交，则没有实数根。

The shape of the parabola also matters: when a > 0, the parabola opens upward and has a minimum point; when a < 0, it opens downward and has a maximum point. The turning point of the parabola is found from the completed square form as described earlier.

抛物线的开口方向也很重要：当 a > 0 时，抛物线开口向上，有最小值点；当 a < 0 时，抛物线开口向下，有最大值点。抛物线的顶点可通过前面介绍的配方法形式求得。

7. Word Problems Involving Quadratics | 二次方程应用题

Many IGCSE questions present quadratic equations in real-world contexts, such as geometry, projectile motion, or number problems. The key is to translate the words into an equation, solve it, and then check whether the answers make sense in context.

许多 IGCSE 题目将二次方程置于实际情境中，如几何问题、抛体运动或数字问题。关键在于将文字转化为方程，求解后再检查答案是否在情境中合理。

For example: The length of a rectangle is 3 cm more than its width, and its area is 28 cm². Let the width be x cm. Then the length is (x + 3) cm, and the area gives x(x + 3) = 28, so x² + 3x – 28 = 0. Factoring gives (x + 7)(x – 4) = 0, so x = -7 or x = 4. Since width cannot be negative, x = 4 cm and the length is 7 cm.

例如：一个长方形的长比宽多 3 cm，面积为 28 cm²。设宽为 x cm，则长为 (x + 3) cm，面积给出 x(x + 3) = 28，即 x² + 3x – 28 = 0。因式分解得 (x + 7)(x – 4) = 0，因此 x = -7 或 x = 4。由于宽度不能为负，x = 4 cm，长为 7 cm。

Always remember to discard any solution that violates the conditions of the problem, such as negative lengths, negative time, or other impossible values. This step is crucial for earning full marks in word problem questions.

务必舍弃任何违反题目条件的解，例如负长度、负时间或其他不可能的值。这一步对于在应用题中得满分至关重要。

8. Common Mistakes and How to Avoid Them | 常见错误及避免方法

Students often make predictable errors when solving quadratic equations. Recognising these mistakes in advance can significantly improve your accuracy in examinations.

学生在解二次方程时常犯一些可预见的错误。提前识别这些错误可以显著提高你在考试中的准确率。

Forgetting to rearrange the equation to standard form ax² + bx + c = 0 before applying methods.

忘记先将方程整理为标准形式 ax² + bx + c = 0 再使用各种方法。
Dividing both sides by x when x is a possible solution, which loses the root x = 0.

当 x 可能是解时，两边同时除以 x，从而丢失 x = 0 这个根。
Making sign errors when substituting negative values into the quadratic formula.

将负值代入求根公式时出现符号错误。
Misapplying the zero product property: (x + 2)(x + 3) = 5 does not mean x + 2 = 5 or x + 3 = 5.

误用零乘积性质：(x + 2)(x + 3) = 5 并不意味着 x + 2 = 5 或 x + 3 = 5。
Forgetting to simplify surds such as √44 to 2√11 when giving exact answers.

在给出精确答案时忘记化简根式，例如将 √44 化简为 2√11。

A good habit is to substitute your solutions back into the original equation to verify them. This takes only a few seconds and can catch careless errors before you move on to the next question.

一个良好的习惯是将你的解代回原方程进行验证。这只需要几秒钟，却能在你进入下一题之前捕捉到粗心造成的错误。

9. Worked Example: Mixed Methods | 综合示例：多种方法

Let us solve the equation 3x² – 5x – 2 = 0 using both factorisation and the quadratic formula, to demonstrate that both methods produce the same answer.

让我们用因式分解法和求根公式两种方法解方程 3x² – 5x – 2 = 0，以证明两种方法得到相同的答案。

Method 1 — Factorisation. We need factors of 3 × (-2) = -6 that add to -5. These are -6 and 1. Rewrite: 3x² – 6x + x – 2 = 0. Group: 3x(x – 2) + 1(x – 2) = 0, so (3x + 1)(x – 2) = 0. Hence x = -⅓ or x = 2.

方法一 — 因式分解。我们需要 3 × (-2) = -6 的因子对，使其和为 -5。这两个数是 -6 和 1。改写：3x² – 6x + x – 2 = 0。分组：3x(x – 2) + 1(x – 2) = 0，即 (3x + 1)(x – 2) = 0。因此 x = -⅓ 或 x = 2。

Method 2 — Quadratic Formula. Here a = 3, b = -5, c = -2, so the discriminant is b² – 4ac = 25 + 24 = 49. Therefore x = (5 ± 7) / 6, giving x = 12/6 = 2 or x = -2/6 = -⅓. Both methods agree perfectly.

方法二 — 求根公式。这里 a = 3，b = -5，c = -2，判别式为 b² – 4ac = 25 + 24 = 49。因此 x = (5 ± 7) / 6，得 x = 12/6 = 2 或 x = -2/6 = -⅓。两种方法完全一致。

The discriminant being a perfect square (49) indicates that the quadratic expression could indeed be factored over the integers. This is a useful quick check before attempting factorisation.

判别式为完全平方数（49）表明该二次表达式确实可以在整数范围内因式分解。这是在尝试因式分解前一个有用的快速判断依据。

10. Exam-Style Practice Questions | 考试风格练习题

Now it is your turn. Try these questions, covering all the methods discussed in this article, and check your answers carefully.

现在轮到你了。尝试以下题目，涵盖本文讨论的所有方法，并仔细核对你的答案。

Question 1: Solve x² – 7x + 12 = 0 by factorisation.	第 1 题：用因式分解法解 x² – 7x + 12 = 0。
Question 2: Use the quadratic formula to solve 2x² + 3x – 5 = 0, giving exact answers.	第 2 题：用求根公式解 2x² + 3x – 5 = 0，给出精确答案。
Question 3: Complete the square for x² – 8x + 15 and hence solve the equation.	第 3 题：对 x² – 8x + 15 配方，并由此解方程。
Question 4: Find the value of k such that the equation kx² + 4x + 1 = 0 has exactly one real root.	第 4 题：求 k 的值，使方程 kx² + 4x + 1 = 0 恰好有一个实数根。
Question 5: A right-angled triangle has hypotenuse 13 cm and one leg 5 cm longer than the other. Find the length of the shorter leg.	第 5 题：一个直角三角形，斜边为 13 cm，一条直角边比另一条长 5 cm。求较短直角边的长度。

Answers: 1) x = 3 or 4. 2) x = 1 or x = -5/2. 3) (x – 4)² – 1 = 0, so x = 3 or 5. 4) For one root, the discriminant must be zero: 16 – 4k = 0, so k = 4. 5) Let the shorter leg be x cm; x² + (x + 5)² = 169, giving 2x² + 10x – 144 = 0, so x² + 5x – 72 = 0; hence x = 8 cm.

参考答案：1) x = 3 或 4。2) x = 1 或 x = -5/2。3) (x – 4)² – 1 = 0，所以 x = 3 或 5。4) 有一个根时判别式为零：16 – 4k = 0，所以 k = 4。5) 设较短直角边为 x cm；x² + (x + 5)² = 169，得 2x² + 10x – 144 = 0，即 x² + 5x – 72 = 0；故 x = 8 cm。

11. Summary and Revision Tips | 总结与复习建议

Solving quadratic equations is a fundamental skill that appears in many areas of IGCSE Mathematics. There are three algebraic methods — factorisation, the quadratic formula, and completing the square — and each has its own strengths.

解二次方程是 IGCSE 数学中许多领域都会用到的基础技能。有三种代数方法——因式分解、求根公式和配方法——每种方法各有优势。

Use factorisation when the expression factors neatly, especially when the discriminant is a perfect square.

当表达式可以整洁分解时使用因式分解法，尤其是判别式为完全平方数时。
Use the quadratic formula when factorisation is difficult or impossible; it always works.

当因式分解困难或不可能时使用求根公式；它始终有效。
Use completing the square when you need the turning point of a parabola or when deriving results.

当你需要抛物线的顶点或推导结论时使用配方法。

Finally, always check the discriminant first when you are uncertain about the number of solutions, and always verify your final answers by substitution. Regular practice with past paper questions is the most effective way to build confidence and speed.

最后，当你不确定解的个数时，先检查判别式；始终通过代入验证你的最终答案。定期练习历年真题是建立信心和提高速度的最有效方法。

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Solving Quadratic Equations | 解二次方程